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What are the areas of an aircraft that suffers most from G forces?
i want to ask this time as to "what are the areas of an aircraft that feels or suffers the most from G forces?". i am asking this in terms of structural design and physical limits of the pilot(the position and seat angle) and the plane.
please try not to limit the airplane model to simple light-aircraft,try to expand it near to military Aviation and other else.
if possible, please try not to use high-falutin words since i'm having a hard time with words in English.
If you can not understand this clearly, i am sorry for my poor English and Grammar, and also i want to elaborate more, but i can't translate it properly at the moment.

Basically 3 places on an airplane see very high concentrations of stress in a small area from in-flight G loads.  In this case we'll say a single engine light aircraft.

Wing root spar fittings on a cantilever wing - the bottom in tension and the top in compression. For a high wing airplane with struts, it will be the strut attachments at each end that join the strut to the fuselage and wing.
The roots of the horizontal tail, same except upside down - bottom in compression and top in tension.
Engine mount attachment to the firewall - top mount in tension and the bottom mount in compression.

Those are the points where flight loads from a lot of mass are concentrated at a single point or set of points during high G loading and are where the structural elements will be heaviest.  For most other areas the loads are distributed over more of a large area.

I think this answer would be spot on if you add “the pilot” to the top of the list.

Especially if he suffers from hemorrhoids!

oh ok, ill change it a little bit, so that people who want to give answer might understand it more

Wing attachment points. (spars and struts)
Not every answer requires paragraphs of postulating and philosophizing.
Reference: 115 years of heavier-than-air aviation accident reports.
In 2002 all large air tankers were grounded after two of them had wing spar failures a month apart. Both were large, 4-engine ex-military airplanes.
Five days ago the FAA proposed an AD on Piper wing spars because the wings keep falling off.
Lowell Bayles was killed in a Gee Bee racer in 1931 when the wing spar failed and the wing ripped off.
In 2012 the EASA issued an AD on the A380 because of cracking in the wing spars after they “found cracks in almost all of the planes inspected”.
This is a straightforward question with a simple answer. All airplane types and sizes have the same major mode of structural failure due to stresses imposed by the force known as "Gs".

Well I think it’s an answer, just not a good one. I’d recommend elaborating and adding some references if possible.

It is THE answer. There is the supporting structure (the wing), and the thing being supported (the fuselage). The most susceptible area is where these two are joined, which is the spars and supporting struts. Not every answer requires paragraphs of postulating and philosophizing.
Reference: 115 years of heavier-than-air aviation accident reports.
In 2002 all large air tankers were grounded after two of them had wing spar failures a month apart. Both were large, 4-engine ex-military airplanes.
Five days ago the FAA proposed an AD on Piper wing spars because the wings keep falling off.

Lowell Bayles was killed in a Gee Bee racer in 1931 when the wing spar failed and the wing ripped off.
In 2012 the EASA issued an AD on the A380 because of cracking in the wing spars after they “found cracks in almost all of the planes inspected”.
This is a straightforward question with a simple answer. All airplane types and sizes have the same major mode of structural failure due to stresses imposed by the force known as "Gs".

G forces are forces caused by the acceleration of a mass. The one acceleration all parts are subjected to all the time is gravitational acceleration, and here John K's answer gives the right direction.
However, there are other accelerations at work which might cause higher loads still. Any fast, cyclic movement will cause accelerations which can become higher than those of gravity, especially when the acceleration is vertical so gravity comes on top. Obvious candidates are:

Flutter: Here, certification requires the mass balance of control surfaces to withstand 24g (JAR 23.659). That does not mean that in every flutter case this acceleration is reached, but that there have been cases where a stronger mounting of the balance mass could have prevented an accident. In case of flutter, the concentrated mass of the mass balance could be ripped off if not adequately secured to the control surface, rendering it ineffective.
Imbalance: This could be the rotating part itself (propeller, engine) or some structural member which is excited at one of its eigenfrequencies. Especially in helicopters, some vibration and shaking is always present and will create local stresses which can easily surpass those of gravity if a part is not well balanced. Repeated, cyclic stresses will let the vibrating part accumulate creeping damage until it breaks suddenly and unexpectedly.

Excuse my use of technical terms - please copy and paste the answer into Google Translate if you need help.


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Why are airspeed gauges labeled above Vne?
Examples:

Cessna 150L Vne = 141 knots (162 MPH), gauge labeled to 170 knots. It looks like there's room to extend the scale to 180.
Cessna 152 Vne = 149 KIAS, gauge labeled to 170.
Cessna 172R Vne = 163 KIAS, gauge to 200

Is the extra range deliberate? Is there a use for it?
Is this specific to Cessna models? I browsed instrument panel photos and didn't find a non-Cessna example. For most aircraft, it seems the range extends past Vne only to the next convenient round number.

I can think of several reasons, why you would want to have some extra room:

The fact that you should not exceed $ V_\mathrm{ne} $, does not mean you never actually do. It would be good to know by how much you are currently exceeding it, so you can decide how to react to this situation.
Not every aircraft model gets their own specific gauge. Cessna probably has a few gauges available and put the one with the smallest extra room above $ V_\mathrm{ne} $ on the aircraft. This is quite similar to cars, where you get the same speedometer on most models of a car, despite having different engines and thus different maximum speeds. It is simply cheaper to produce (and certify) only a few different airspeed indicators.
The aircraft's $ V_\mathrm{ne} $ might change with modifications installed at a later time (like an aerodynamic upgrade package). It would be nice, if the avionics would not need to be exchanged for this.

As to whether or not it is specific to Cessna: A quick Google search for "PA28 airspeed indicator" shows several Pipers with a lot of extra room (although some only have around 10 knots). I assume this is not much different for other aircraft manufacturers and just depends on which gauges where available.

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Do you need clearance to fly VFR into a Warning Area
PHAK is not clear about VFR flying into Warning Areas. Warning areas are similar in nature to restricted areas, which says IFR can be cleared through the airspace but VFR is not mentionned.
Sectional uses the same visual cue as Prohibited and Restricted but that's a little thin to draw conclusions.
AIM says Prohibited and Restricted are ruled by 14 CFR Part 73 but nothing seems to rule Warning Areas: Warning areas, military operations areas (MOAs), alert areas, and controlled firing areas (CFAs) are nonregulatory special use airspace.
Do you need clearance to fly VFR into a Warning Area?

Related: What are some examples of warning areas and restricted areas?

No, unless explicitly directed by ATC not to enter a warning area, you can enter one while flying under VFR.  Most warning areas are located outside of the air defense identification zone (ADIZ) which would require filing a flight plan and obtaining a clearance to cross.  Once out there, you’re in international waters, and not under the jurisdiction of the FAA.
That being said, if you intend to enter a warning area on a flight, it would be wise to obtain a preflight briefing and inquire if the warning area will be active during the time of your intended flight. In addition, if ATC is warding people away from a warning area at the behest of the military, it will be very wise to heed their advice and stay out of there.  Fools rush in where angels fear to tread…

As someone who used to schedule missions in warning areas a long time ago, I second Carlo's recommendation: If the area is hot, stay out. If they don't see you, things could go very bad. If they do see you, they'll likely abort the mission. That ends up wasting the day for a whole bunch of people. Don't be that guy.

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Does ATC use True Course or Ground Track?
Reading the practice questions for PAR, I came accross this

ATC advises, "traffic 12 o`clock." This advisory is relative to your
  A) true course
  B) ground track
  C) magnetic heading

I was inclined to got for A) but isn't that the same as B)?
Neither are defined in PHAK glossary.

A heading is the direction the plane's nose is pointed. A course is the direction it is actually traveling. The difference between the two depends on wind.
A true heading or course is corrected for magnetic variation; a magnetic heading or course is not.
Track and course are often used interchangeably, but technically a "course" refers to what you intend to do while a "track" refers to what you actually do.
ATC can only see your ground track on their radar screens, so logically that would be their reference for traffic advisories--and you may look in the wrong direction if you don't realize that, which I suspect was the point of the question.

Refer to AIM section 4-1-15. There's even some nice diagrams in there. "Since the radar controller can only observe aircraft track (course) on the radar display, traffic advisories are issued accordingly."
Answer (A) is incorrect because ATC course/heading/azimuth information is always in magnetic, not true.

Thanks, The PDF is daunting at first, but manageable when you cherry pick your content.

because ATC course/heading/azimuth information is always in magnetic
I think whether it's magnetic or true doesn't really matter here, as the ATC is not announcing xx degrees.


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Why does a hovercraft lift blade look one way, but a propeller another?
The lift blade of a hovercraft looks like this:

Note the ten rectangular blades.
The blade of a propeller-craft looks this this:

Note the three, traditional airfoil-shaped blades.
Why are the two different?
Edit: My question is different from the other question cited because it asks about both the number of blades and their shape, as well as other differences I may as of yet be unaware of.

Possible duplicate of Why don't aeroplanes use propellers with more blades like the intake on a fanjet?

In my opinion, this question isn't a duplicate of "Why don't aeroplanes use propellers with more blades like the intake on a fanjet?", because the answers on that question don't address the question of why hovercraft lift blades are designed the way they are.

@TannerSwett: Duplicates should be judged by the questions content (answers may be incomplete)

Because lift propellors need to provide static pressure to inflate the airbag and hover. This means more fan blades to stop backflow.
The thrust propellors need to provide force which is a combination of static pressure and airflow but relies mostly on airflow. This means less fan blades to facilitate large airflow.

But turbofans provide a very large airflow and have a lot of blades...

@xxavier but their diameter is constrained and the duct behind them is longer so they benefit more from static pressure.

There are two parts to this: number of blades and squared off tips.
1. Number of blades.
A propeller blade is actually just a wing. This is true regardless of weather you are talking about a piston engine airplane, a jet engine, a table fan or the propeller of a cargo ship.
A wing is most efficient if the air coming onto the wing is undisturbed. While it is possible to design wings that work efficiently in very turbulent air (for example insect wings, in particular google about research on dragonfly wings) we are not quite there yet in terms of either theory or technology.
As a consequence, adding more blades to a propeller means each "wing" gets interference from the blade in front of it. More blades decreases efficiency. The most efficient prop of all is the single bladed prop. Some motorised glider designs actually use single bladed props that can be driven with tiny lawnmower engines.
However, a single blade leaves a lot of air behind it that can potentially move back up to the low pressure region just created by the prop. Thus, a single bladed prop may be very efficient (uses less fuel to travel the same distance) but generates less thrust. The more blades you add the more thrust the propeller can maintain.
If you fly drones you can see this in drone design. Larger drones use two bladed props because their size means that the props generate sufficient thrust and using two bladed props improves battery life. Smaller drones cannot use large props but must generate thrust to hover. Therefore smaller drones are designed with three or four bladed props.
For a hovercraft, the hovercraft is not designed to travel upwards forever. They are designed to generate just enough lift to hover. This means the lift prop never gets unloaded (the effect of props gaining efficiency when operating in a moving air stream). So for a hovercraft lift prop generating high thrust while not moving off the ground is paramount. Adding more blades allow you to generate more thrust with the same engine (though it will burn more fuel).
It is possible to get the same amount of thrust with a two bladed prop - just make the prop bigger. But a hovercraft is size constrained - you want the prop to be fully contained within the fuselage (this is also why jet engines need more blades).
2. Squared off tips.
Admittedly this is a secondary and less important factor since some aircraft propellers also have squared off tips - but you did ask.
As mentioned above a blade is just a wing. One of the problems with wing design is the loss of lift and increase of drag due to tip vortex. One of the solutions engineers have found is to make the lift distribution from wing root to wing tip to be as close to elliptical as possible. This can be done literally by making the wing elliptical (like the Spitfire) or by altering the wing incidence to be less at the tip and more at the root, basically twisting the wing (often called a washout). If you can't get a perfect elliptical distribution then a rounded tip would still improve lift generation and drag.
However, tip vortex cannot form on a hovercraft lift blade because the blade is rotating inside a shroud or duct. The duct physically blocks air from curling around the blade tips. As such you can square off the blade without losing performance. In fact for optimum performance you'd want zero gap between the blade tips and the duct therefore a squared off tip that just fits the duct is better.


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Is separation provided in class F airspace?
According to the Wikipedia page on different airspace classes, separation is provided for IFR/SVFR to other IFR/SVFR in class F airspace. In ICAO Annex 11 from 2018 (p.36 in the pdf file), I can't find anything about separation being mentioned.

Class F. IFR and VFR flights are permitted, all participating IFR flights receive an air traffic advisory service and all flights receive flight information service if requested.

I'd like to know if I have missed something and the Wikipedia page is right nevertheless?


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What's the relationship between aileron deflection and roll moment?
I'm wanting to build a laterals-only kinematic model in MATLAB considering only the ailerons for control inputs and was wondering if anyone has a rough guideline for the relationship between their deflection and the resulting roll moment.
EDIT: The answer from @PeterKämpf on the following question helped a lot (Thankyou Peter) but I couldn't understand the difference between p, dimensionless roll rate and wx, rolling speed in rad/sec. How can a roll rate be dimensionless?
His answer is here: How I determine numerically the roll rate of an aircraft?
Sorry, I couldn't comment on the actual post, I don't have enough reputation.

Related? How to calculate aileron deflection for achieving similar roll dynamics at thin and dense air?

The lateral and directional modes are usually coupled, so much so that the usual distinction is between longitudinal and lateral-directional stability. I trust you are aware?

@AEhere I wasn't, I have only learnt about longitudinal and lateral dynamics, I didn't know there was a third to consider. Thank you I'll take a look.

@Rim3rd this might be a translation issue, but I mean the yaw-roll coupling. In most cases your aileron-only input will cause adverse yaw, which will feed back into your roll response.

@AEhere Ah I see, yes thank you, I was aware. I'm only looking for an equation relating aileron angle with roll rate at this stage.

@Rim3rd you can get one for the torque, but not for the roll rate unless you are willing to assume the aircraft is constrained in yaw.

@AEhere OK I think I understand... Is it a better question to ask for a relationship between aileron angle and moment about the body x axis?

@Rim3rd it would certainly make it simpler... I'll try to pen an answer later today when I can look up the formulae, I don't trust myself not to forget a term somewhere.

@AEhere Most kind, thank you.

Using a linear theory, the roll moment coefficient can be expressed as:
$$
C_l =C_{l0}+C_{l\beta}\beta+C_{l\delta_{\alpha}}\delta_{\alpha}+C_{l\delta_{r}}\delta_{r}
$$
, where:
$\beta$ : slip angle
$\delta_{\alpha}$ : aileron deflection
$\delta_{r}$ : rudder deflection
This doesn´t really say much, as we are simply defining this coefficient in a way that is convenient. This equation just says that $C_l$ is equal to a linear composition of the effects of assymetry, slip angle, and aileron and ruder deflections, each weighed by a stability derivative.
Some of these stability derivatives have proper names or are trivially discarded:

$C_{l0}=(C_l)_{\beta=\delta_{\alpha}=\delta_r=0}=0$ , for symmetric aircraft without engine torques
$C_{l\beta}$ is the dihedral effect, or the effect of slip on roll. It is the sum of contributions from the wing-body assembly, the HTP and the VTP. The wing-body effect can be further split into: the geometric dihedral of the wing, wing sweep and vertical position (high or low).
$C_{l\beta} =(C_{l\beta})_{WB}+(C_{l\beta})_{HTP}+(C_{l\beta})_{VTP}$
$C_{l\beta}=-\frac{a_w \Gamma}{4}-\frac{C_L}{4}sin(2\Lambda)+(C_{l\beta})_{\text{wing height}}+(C_{l\beta})_{HTP}-a_v\eta_v\frac{S_v}{S}(1+\frac{\partial\sigma}{\partial\beta})\frac{h_v}{b}$
$(C_{l\beta})_{HTP}$ is obtained the same way as the WB contribution, although with different adimensionalizers.
For general reference, wingtips above the root are stabilizing, positive sweep is stabilizing and the wing height effect depends on the fuselage fraction ahead and behind the wing, but usually a high wing should be stabilizing.

$C_{l\delta_{\alpha}}$ is the roll control power or the effect of ailerons on roll. Is is calculated directly from the Prandtl wing theory by considering the impact of aileron deflection on the lift distribution
$C_{l\delta_{r}}$ is the effect of rudder on roll
$C_{l\delta_{r}}=C_{Y\delta_r}\frac{h_v}{b}=-a_v\eta_v\frac{S_vh_v}{Sh}\tau_r$

Your original question asked for the roll rate, and Peter Kämpf does indeed provide an answer, but it is probably no more useful to your particular need than my very first equation.
If you are interested in the roll rate, you will need to solve the lateral-directional system. You can simplify it somewhat by allowing fixed controls and level fight, thus obtaining a stability cuartic. This has, for most configurations, two real and two complex conjugate solutions,  corresponding to the three lateral-directional eigenmodes: Roll Subsidience (large negative real root), Spiral Divergence (small real root, sometimes even >0) and Dutch Roll (complex conjugate root). Since the large negative real eigenvalue has a much greater modulus than the rest, its mode (Roll Subsidience) will dominate for short periods, and you can obtain a good approximation of the roll transfer function by doing $\Delta\beta=\Delta\hat{r}=\Delta\delta_r=0$ and $C_{l\hat{\dot{\delta}}_\alpha}\cong0$:
$$
G_{\hat{p}\delta_\alpha}=\frac{\Delta\hat{p}(s)}{\Delta\delta_\alpha(s)}=\frac{C_{l\delta_\alpha}}{\hat{I_x}}\frac{1}{s-(C_{l\hat{p}}/\hat{I_x})}
$$
...in the Laplace plane. Keep in mind the parameters are non-dimensional.

Thank you for taking the time to help me with such a detailed response! You've given me a lot to read up on and it's much appreciated. I have voted up your answer but apparently I don't have the reputation for it to display. Thanks again!

@Rim3rd No need, it was a welcome reminder for myself as well. I used one of my professor's books on the subject, which afaik has not been translated into English, but you can get a good, if simpler, overview from this resource: https://courses.cit.cornell.edu/mae5070/DynamicStability.pdf

This is a great resource! Thanks very much!


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Calculation QNE and QNH, with flight level and QNH set lower (higher) than QNE
Hi Student pilot here.
Lets say Mr pilot is flying at FL75 and QNH is set at 995 hpa, however there is a mountain in the way at 7000', will he below or above the mountain?

If you could visually explain that would be even better.

Since you used the term "Flight Level" (FL75), we must assume that the altimeter is set to 1013.25, and that the altimeter reads 7500'.
The math is basically this: each milibar (or hPa) is worth about 27 feet.  We know that his altimeter is set to 1013 hPa, but the actual QNH is 995 hPa, a difference of 18.
18 * 27 = 486.  So, the aircraft is either 486 feet ABOVE 7500, or 486 feet BELOW 7500.
So.... how do you know whether to add or subtract?  We can remember the phrase "High to low, look out below"... but what does that really mean?
Let's say that an aircraft is in an area where the QNH is really 1013, and is at an altitude of 7500'.  The pilot flies along the route for a while without adjusting his altimeter. The QNH has been steadily dropping (unbeknownst to the pilot) and is now 995 hPa. Since the pilot failed to obtain the local QNH as he passed by various stations (or, by asking Air Traffic Control every hour or so), he has also failed to adjust his altimeter to the new QNH reading.
Here is what has happened: As the aircraft continues to fly from the higher pressure area into a lower pressure area ("High to low"), the altimeter will start to go up. The pilot will unknowingly descend in an attempt to keep the aircraft at the desired altitude of 7500'. The aircraft will continue to get lower and lower ("look out below"), but all the while the pilot thinks he is maintaining straight and level.
So, we know that the aircraft is 486 feet BELOW his indicated altitude.  Therefore, the aircraft is actually at an altitude of 7014'.  He will BARELY miss that mountain peak that is at 7000'. If the QNH was one milibar lower, he would be at 6987'and hit the mountain.

Let's hope there are no trees at the top of the mountain, lest Mr Pilot might clear the mountain proper but still end up with the aircraft in a rather undesirable state.

Yeah, lots of things wrong with this scenario, including the use of the term “Flight Level” to mean altitude. There would not be a FL75 in mountainous terrain. But, the exercise was to calculate the math, so I hope I helped the OP understand it better.

Much appreciated Jimmy! this helped me allot. Just finished my initial batch of 9 tests (AGK,POF,FPP,NAV,COM,HUM,OPS,LAW,MET) passed them all, and just completed my 7th flight.

It should be noted that in practice the transition level will be high enough (FL100+ in this case) to prevent this scenario from happening, so the question may be intended to illustrate why that's needed.


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Identify a change in Altitude based on direction and WCA applied?
Question below is from study material for the EASA PPL license MET course.
It's an unlikely question but it's interesting to review.
You are flying from west to east with WCA +6°. How is your true altitude changing with all else remaining un changed.

A) It will degrees
B) It will increase
C) Will remain stable(unchanged)
D) None of the above are correct.

My Question, how is this easily explained, does it have to do with Buys Ballot's law ?

Welcome to the site! It's always a good idea to ask a specific question and show your work before posting. Are you having trouble answering this? If so, what do you need help with?

Hi GdD ;), Agree i´m not very clear!. The answer is i think (B) it will increase i.e Buys Ballot comes into play. Maybe i´m not getting this out correctly, but what is the simplest way to understand this.

The question is confusing because the graphic doesn't match the question being asked.  Does the question actually use this graphic?  Because if it does then your WCA would be negative if you are heading East because you would be crabbing to the left to account for wind coming from the North.
In any case, if you are flying from an area of high pressure to an area of low pressure (as the graphic depicts) without changing your altimeter setting your true altitude will be lower.  Remember this simple mnemonic:  "High to low, look out below."
However, if the question is as stated WITHOUT the graphic, then heading East with a positive WCA means that the wind is from the South.  If this is the case it would be fair to assume from Buy Ballot's law that you are transiting from an area of low pressure to an area of high pressure.  In this case then the answer would be your true altitude would be higher.

Thank you Michael, question does not include the picture. But your explanation has help me ;)


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What is the minimum speed at which the Antonov An-2 can maintain level flight?
The An-2 famously has "no stall speed", as control can be maintained in descent at arbitrarily low airspeed. As wikipedia says:

The An-2 has no stall speed, a fact which is quoted in the operating handbook. A note from the pilot's handbook reads: "If the engine quits in instrument conditions or at night, the pilot should pull the control column full aft and keep the wings level. The leading-edge slats will snap out at about 64 km/h (40 mph) and when the airplane slows to a forward speed of about 40 km/h (25 mph), the airplane will sink at about a parachute descent rate until the aircraft hits the ground."[4] As such, pilots of the An-2 have stated that they are capable of flying the aircraft in full control at 48 km/h (30 mph) (as a contrast, a modern Cessna four-seater light aircraft has a stall speed of around 80 km/h (50 mph)). This slow stall speed makes it possible for the aircraft to fly backwards relative to the ground: if the aircraft is pointed into a headwind of roughly 56 km/h (35 mph), it will travel backwards at 8.0 km/h (5 mph) whilst under full control.

But there has to be some minimum speed at which it can generate just enough lift to counteract gravity, what is this speed (and does it have a general name)?

Don't know the value for an AN-2 , but the speed you are looking for is usually reffered as Vs1g

"No stall speed" just means no stall break in the traditional sense.  The AN-2's slatted wing's maximum AOA is very high (typically about 10+ degrees higher than un-slatted, or mid-high 20s vs mid teens) and there isn't enough tail power to get the AOA high enough to get any kind of break.
Stall speed is normally published as the speed at which the wing is at maximum lift coefficient.  A typical wing makes a Clmax of about 1.5, slats raise that to about 2.5, and a slotted flap raises that again to about 3.5-ish. The AN-2 has full span slats and flap-flaperons so you can assume that just about all of its 770 sqf of wing area is producing a Clmax of mid 3s with slats out and flaps down.
You can use the handy calculator on this page (click on the "Stall Speed Calculator" link) to work out what the speed is at which this is achieved. For the AN-2 you could assume with 770 sqf of wing area, with, for argument's sake, a Clmax of 3.5 for the entire wing, and operating at, say, a relatively light 9000lbs, you get a "stall" speed of 32.8 kts (37 mph), the speed that the wing is making maximum lift based on those assumptions and technically, this is the answer to your question.
With a slatted wing, the lift curve has a more rounded top; this is, lift declines fairly gently after the peak, so you can continue raising AOA and all that happens is sink rate goes up, until the tail runs out of downforce authority.
With power, thanks to the slats, you get both more tail downforce with the slipstream blowing on it, and because you can reach very high deck angles, you start to get a significant boost from the vertical component of engine thrust itself when the thrust line is canted up at, say 25 degrees.  And on top of that you get added lift from the slipstream itself passing the wings..
If the engine is making, for argument's sake, 4000 lbs of thrust when at full power (you get roughly 4 lbs/hp) and the thrust line is canted up 25 degrees, there is a vertical thrust component of about 1700 lbs.  This has the effect of reducing your weight by 1700 lbs, and plugging in a 1700 lb weight reduction into the calculator gets you down to about 29 kts (or 33 mph).  Add in the additional lift from slipstream passing part of the wings dropping the minimum speed a few more mph, and you can see how you can get to the point of the plane being able to back up at 5 mph in a 35 mph headwind using power in the slow-flight regime, at a relatively light weight at least.
So, the AN-2, at a relatively light 9000 lbs, using the standard formula and using typical assumptions, should be able to fly at a minimum speed of 37 mph power off and about 30 mph or a little less power on without sinking, which agrees more or less with the article's statements.
Plus, there is enough tail power with power on to increase AOA a little more, getting speed down to 25 mph, but while it's under control, total lift will be declining and the airplane will be sinking while doing so.
None of this is particularly special.  The AN-2 achieves its performance with lots of wing area and full span slats and flaperons.  Other STOL aircraft with full span flaps and slats can do the same thing.

Generally speaking, the stall speed doesn't exist as the actual speed at which you might stall a wing depend on the loading of the wing. In a 60° level turn you'll experience a force of 2g's and the stall speed of any given aircraft will be increased by roughly 1.5. What is important is the stall angle of attack (AOA) which is defined by the profile of the wing. Keep in mind that lift doesn't vanish when you stall a wing but rather it's the drag vs lift coefficient that increase significantly.
I don't know much about the AN-2 but my explanation of what seems to be described in the Wikipedia quote would be a lack of elevator authority. This means that the elevator cannot pitch the nose up significantly to increase the AOA of the main wing beyond its stall limit. With the engine out, you can reach an equilibrium point where for a given speed (here 40km/h) your elevator generates just enough force to pitch the nose up to a given AOA. And let's say that at this AOA none of the surface are stalled. If your speed decreases the elevator will lose efficiency and the nose will pitch down, thus increasing your speed, giving you more elevator authority, pitching the nose up again, losing speed .... and so on. You'll never stall the aircraft because you can't pass beyond the critical AOA and thus you can keep aileron authority and "fly" the aircraft. But there are no way to ease your landing so this is pretty much the same situation as pulling the emergency chute of your Cirrus in case of the engine dies out in IMC conditions.
"Control can be maintained in descent at arbitrarily low airspeed." Following the above explanation I'm actually not sure you will be able to slow the aircraft below 40km/h so you don't have arbitrarily low airspeed.
For the second part of your question there are 4 forces interacting in flight. Thrust, gravity, lift and drag. To fly level, you need to balance out everything.
The lift equation can be written as $C_l*AOA*V^2$ where Cl is a constant with air density, surface area and lift coefficient of the wing. Depending on the AOA the speed required to have lift=gravity may vary. And you will still have to balance drag with thrust. And at high AOA, thrust will have a non-negligible vertical component that will be added to the lift to balance out gravity. Thus your minimal speed required to fly level will also depend on your engine power. If thrust to weight ratio is greater than 1, the speed required to fly level can be 0.


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How to take into account man hours in a Service Bulletin?
There are two phrases in any SB included "Total Man hour"and"Elapse time", which quantity should we use in job card for accomplishment an AD or SB?

Elapsed time is the total man hours divided by the number of techs assigned to the task.  So 40 manhours will take one tech 40 hours to complete or 2 techs 20 hours.  If the number of techs to be assigned is up to the supervisor or a scheduler/planner, I would just include the total man hour estimate.  If the job card specifies the number of techs required for the job, then include the elapsed time based on that.
Note that on the first several aircraft modified under an SB, the actual man hours will exceed the SB estimate by a large margin, maybe 50% or more, until the technicians involved get through the learning curve of the task, and even then will probably never get to the SB labour estimate, which assumes perfect conditions and no wasted time.  If the OEM is giving a labour credit that can be charged back to the OEM at some hourly rate, it will be based on the SB estimated labour, which is derived from the proof installation of the SB, not the actual hours used.


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How is the time of "threshold" and "phase in" in the Airworthiness Directive?
For example Within the next [25 hours] time-in-service (TIS) after the effective date of this AD, unless already done within the last [75 hours] TIS, and thereafter at intervals not to exceed [100 hours] TIS from the last inspection.

I know, so I edited the previous question yesterday, but I don't receive any answer yet

Is there any guidance about analyzing of ADs & SBs that help me?

The threshold is the starting base line of the repeat inspection requirement, in this case the calendar date of AD release, and the phase-in is the "grace period" (25 hours) to accommodate airplanes not done yet.
The AD implements a hundred hour repeat inspection.  That means that there is a maximum dormancy (potential undetected existence of whatever condition it is) of 100 hours from the last time the condition was confirmed not to exist by that specific inspection.  The probability of the condition growing from a detectable level to a dangerous level is acceptably low within 100 hours based on analysis or test.
For airplanes where the condition has never been confirmed not to exist and is unknown, or at least more than 100 hours ago, there is obviously a higher risk profile, and they are not comfortable with waiting more than a full 100 hours of the repeat interval with unknown aircraft. However, they are comfortable with letting unknown aircraft go another 25 hours before the inspection, to avoid grounding everybody until the inspection is done as soon as the AD is released.
So the language says:
You have to inspect at 100 hour intervals going forward, period.
If you've never done the inspection, you can't wait 100 hours and have to do it within the next 25 hours.
If you've already done the inspection for whatever reason before this AD came out, you get credit for that if it meets the maximum 100 hour dormancy.  This is where the 75 hours before the AD comes in.
Example:
Airplane has 1000 hours when AD is released.  Inspection was done 50 hours ago, at 950 total hours, for whatever reason (maybe it was already in place but with a longer interval before, or was a voluntary Service Bulletin, something like that).  It's within 75 hours of the AD release so it can be treated as the first inspection of the 100 hour repeats going forward and you can ignore the 25 hour phase in and do the next inspection at 1050.
If the same 1000 hour airplane had the inspection done at 90 hours ago when the AD came out, at 910 hours, you aren't forced to do the repeat in the next 10 hours (the next 100 hour interval), at 1010; you get the benefit of the phase in and can wait up to 25 hours of AD release before doing it, then every 100 hours going forward.

Is there any guidance that helps me about analyzing of ADs

Not that I'm aware of but you should try googling phrases like "analysis of airworthiness directive compliance" or "logic of airworthiness directive compliance" or something similar.  Once you get a handle on the logic they become fairly easy to figure out, but occasionally the compliance requirements are really complex due to the need to balance risk mitigation vs operational viability for an issue with many different aspects.  Those you just have to kind of map them out and figure out the logic bit by bit.

actually, I want to be an expert about analyzing ADs & SBs, what should I do?

Go on the FAA website and download a range of ADs and figure them out with someone local who has experience with them.  SBs you have to get from the manufacturer or an airline.


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What's the difference between 'modification' and 'corrective action' in an AD or SB?
For example Modification of an airplane in accordance with the instructions of the specific SB as specified in Table 1 of this AD, as applicable...

It would probably help a lot if you could provide examples of each.

such as Terminating Action:
(5) An accomplishment of corrective action(s) on an airplane, as required by of this AD, does not constitute terminating action for the repetitive inspections.....                                                                                (6) Modification of an airplane in accordance with the instructions of the Airbus SB as specified...

[Edit] that into your original post. Additionally, linking to the specific AD or SB you're referring to would be very helpful.

such as AD No. 2017-0051,page 3

I couldn't find an AD 2017-0051 at the FAA's AD website. If you have a link, please [edit] it into your OP.

on the EASA website.

Then you shouldn't have tagged it [tag:faa-regulations]

A modification is a change to the airplane's physical configuration from the standpoint of fit/form/function.  Think Service Bulletin with a parts kit. Or an SB that has instructions for a physical rework of a part that makes it different from non-reworked parts, and the part is re-identified.
A corrective action is any action required by the AD to mitigate the risk associated with it.  Usually it's something less than a mod, like a inspection, change to a maintenance program, crew procedural change, de-escalation of a check or procedure's interval, an operational limitation, or any other action as required by the AD.
Terminating action is some activity that completely resolves the AD and restores the airplane's original airworthiness state, as opposed to something that temporarily mitigates risk so you keep on flying.  Incorporation of a Mod, or permanent incorporation of a inspection regime that previously wasn't mandatory.
On a side note, one thing about FAA ADs that makes them a total pain from the OEM's standpoint is that the FAA will not issue revisions to ADs like other authorities do.  If there is a change to the AD's terms they will only issue a new AD that supercedes the old one. Makes getting minor changes to ADs very difficult because it has to be worth issuing a new one.


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Does an AD terminate by issuing new one?
When an AD superseded by a new one, Does the operator still retain the instruction of superseded AD even though the AD include repetitive action?

I'm not completely sure what you're asking. Are you asking if a new version of an AD always supersedes any previous versions?

That sounds like a typical US AD where as I mentioned in a previous post, they don't issue "revisions", like AD 2019-12, AD 2019-12 Rev A, Rev B, Rev C etc.   If the AD is basically the same but has some internal change, the FAA will not issue a "Rev A".  They will issue a superceding AD with a new number that is the same as the old one except for various changes which will be spelled out.  So AD 2019-23 will be replaced by AD 2019-32, which supercedes 2019-23, instead of 2010-23 Rev A, as would be done in Canada or elsewhere.
That language tells you that the new AD is issued with administrative changes and that the compliance requirements are the same as the superceded AD, including any repetitive instruction.  In fact the new AD should have all the compliance requirements of the old one included in it.

Mr. John K, thank you so much for helping, then we must close the status of the old rev. AD(repetitive) for applicable A/C and add new replacement AD to the list and comply it.

Yes the old one is superceded so it is no longer in effect.

But  we've bought an A/C and I've seen an AD is open in the AD status that complied with in the previous operator, now that AD is due, such as EASA AD No.2011-0006 that superseded by 2014-0018R1,  Do I take  into account AD 2011-0006 for nearest maintenance check or I ignore it (those ADs are presented in status and 2011-000 6not closed)

The difference between 0006 and 0018 is the requirement for placards with Skydrol resistant glue.  If the airplane was compliant with 0006, the inspection and swap to the one-weld accumulator should have been done and you're covered by the statement "unless accomplished previously" in 0018.  However 0018 does include new placards so you still have to swap out the placard required by 0006 with the better one under 0018.  If 0006 was never complied with at all, then it's like it never existed and just comply with 0018.

Here’s an example of an AD that supersedes an older one. In this case the FAA decided that the requirement that an inspection be performed by an A&P was not necessary and that pilots were smart enough to check whether their fuel selector was labelled correctly. If you find the link to the old one on FAA website, it will show that it has been superseded. But you can still find the old pdf without any indication that it has been cancelled. If you are looking for a particluar AD, make sure you search in the current ADs section of their website. The naming convention differs from the Advisory Circular, FAA Order, and FAA publication method where the old ones are still on the website and the newest one has the highest letter, e.g. AC 61-65E, AC 61-98D, FAA Order JO 7400.9Z.

We are superseding Airworthiness Directive (AD) 2018-02-05 for certain
  Piper Aircraft, Inc. Models PA-28-140, PA-28-150, PA-28-151,
  PA-28-160, PA-28-161, PA-28-180, PA- 28-181, PA-28-236, PA-28-201T,
  PA-28R-180, PA-28R-200, PA-28R-201, PA-28R-201T, PA- 28RT-201, and
  PA-28RT-201T airplanes. AD 2018-02-05 required inspecting the fuel
  tank selector cover to verify the left and right fuel tank selector
  placards are located at the proper positions and replacing those that
  are improperly located with new placards. This AD addresses the same
  unsafe condition and requires the same actions as AD 2018-02-05, but
  changes the inspection of the fuel tank selector cover to a preflight
  check and allows for various fuel selector clocking configurations.


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Could you more explain about the compliance time specifies in the following sentence?
For example Prior to accumulating 5,000 landings or within the next 1,000 landings after the effective date of this AD, whichever occurs later.

I'm not sure what you mean by 'a real-time'. There's no time limit in the wording you've posted, just limits on the number of landings.

Within the next [25 hours] time-in-service (TIS) after the effective date of this AD,
unless already done within the last [75 hours] TIS, and thereafter at intervals not to
exceed [100 hours] TIS from the last inspection, inspect...

It's a threshold and phase-in.  The 5000 landings is the threshold and is landings since new, with a 1000 landing phase-in.
The phase-in is to accommodate airplanes close to, or over, the threshold; otherwise, everybody over the threshold would be grounded immediately, and someone close to the threshold would have only the hours remaining to the threshold to comply.
The effect is, if you have 4000 landings or less on the a/c, you have until 5000 landings since new.  If you have more than 4000 landings, you get 1000 landings from the release date of the AD to comply.
So if you are at 4100 landings, you get to 5100, 4500 landings, you get to 5500.  If you are at 6000 landings, you get to 7000.  And so on.
Just about any AD that doesn't have a hard calendar compliance date, like the one grounding the Max, will have this threshold and phase-in that is negotiated between the OEM and regulator based on risk and probability analysis that establishes how long a problem can be dormant without impacting fleet safety.

I understand thank you so much, but unclearly yet, different between whichever occurs later with whichever occurs first

It's whichever occurs later.  If your airplane has 4950 landings when the AD is issued, you've got 50 landings until you reach 5000.  So 1000 landings landings would occur later and you would have until then to comply.  If you have 1500 landings, you would have 3500 to reach 5000 so that would occur later than 1000 after issue.

All that matters is the "later" part. Think of it this way.  Everybody gets AT LEAST the phase-in period of 1000 landings to comply, but you don't want to penalize someone with a nearly new airplane with a 1000 landing compliance.   So you put in a 5000 landing threshold, so the phase-in, effectively, only applies to airplanes over 4000 landings so they get at least 1000 landings grace period.   People get confused because they see threshold and phase-in and have to realize that the phase-in applies to anybody within the phase-in period short of the threshold, as well as beyond the threshold.

Within the next [25 hours] time-in-service (TIS) after the effective date of this AD, unless already done within the last [75 hours] TIS, and thereafter at intervals not to exceed [100 hours] TIS from the last inspection, inspect... first is the 25 hours threshold? second, if the inspection already is done by the operator means that we only need an inspection at the interval and initial inspection is omitted for this situation

You need to comply with the AD prior to accumulating 5,000 landings or within the next 1,000 landings after the effective date of the AD, whichever occurs later.
This is a very simple order compared to a few others.

actually, my question related to whichever occurs later, could you more explanation about that with example

Within the next [25 hours] time-in-service (TIS) after the effective date of this AD, unless already done within the last [75 hours] TIS, and thereafter at intervals not to exceed [100 hours] TIS from the last inspection, inspect... at first is the 25 hours threshold? second, if the inspection already is done by the operator means that we only need an inspection at the interval and initial inspection is omitted for this situation


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What is the meaning of AFT in airworthiness directives?
What is the meaning of AFT in airworthiness directives?

The average flight time (AFT) is defined as flight hours (FH) divided by flight cycles (FC) accumulated by an individual airplane since the airplane's first flight, specified in hours and hundredths of an hour.

The "special" needs to be done within 34,500 flight cycles or 172,600 flight hours? That's a lot of time in the air, is it not?

it’d be about 20 years for a 6-cycle-a-day short haul aircraft. The test in question seems to be a fairly intensive test for cracks, so presumably it’s something that applies only to old aircraft that the owners want to keep flying.

exactly, A/C was manufactured in 2002


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Advance ratio formula for drone propeller in vertical climb
According to this wikipedia article. There are two expressions for advance ratio:

For propellers:
$$J = {V_\infty \over nD} \tag{1}$$
For helicopters:

$$\mu = {V_\infty \over \Omega r} \tag{2}$$
We know that: $\Omega = 2\pi n$, so by substitution in (2):
$$\mu = {V_\infty \over {2\pi nr}}= {V_\infty \over \pi n D} = {J \over \pi}$$
So $\mu \neq J$
My question:
In case of a helicopter in a vertical climb (no forward velocity) with constant velocity $\vec v = (0, 0, u_z)$, which expression should I use for advance ratio? $J$ or $\mu$?

The advance ratio, since it is not dimensional, is just an intermediate term on your way to something else. For example, for a prop you'd have a chart of coefficient of thrust versus advance ratio. You'd just scale the chart appropriately based on which ratio you were using.

I could not find equation (2) in my copy of J. Gordon Leishman' Principles of Helicopter Aerodynamics. Some relevant info on momentum analysis in forward flight can be found here on page 63.
Advance ratio as a useful measure of blade twist is less meaningful in helicopters, due to the much greater variety in the direction of the velocity vector. It makes more sense to optimise blade twist for forward flight than it is to optimise for vertical flight, a state with much higher fuel consumption which helicopters do not particularly like.
If you would like to use the advance ratio for helicopters as stated in the wiki article, it makes more sense to use the helicopter one. There may be a direct parallel between the states you describe in your question, but for the helicopter case it is not the design state while for the fixed wing it is.
Whether either is relevant to a propellor drone depends on the design of the drone, whether the propellor is shrouded etc. But for a drone, free velocity would usually not be close to perpendicular to the propellor plane, just like in a helicopter.

Thank you, Yes Before posting the question, I read that book but in climb analysis he didn't used the advance ratio. I am only intersted in climb flight.

The advance ration for propellers and helicopters as defined in wikipedia are expressing the same concept, the propellers formula uses "revolutions/second" and helicopters uses "radians/second"


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Comparison of pitching moment at different locations
I've conducted some numerical simulations of a pitching airfoil at three different locations of the pitching axis along the airfoil. The locations (from the leading edge of the airfoil) are: $0.25c$, $0.4c$, and $0.75c$. where  $c$ is the chord length of the airfoil.

Each simulation computes the pitching moment coefficient using the pitching axis itself as the reference point. That is, In, the first simulation the pitching moment is computed about $0.25c$ axis, and in the second simulation the pitching moment coefficient is calculated about $0.4c$, etc.
My questions are:

Is it OK to compare all three moment coefficient together given that the reference points are different?
Do I have to change the reference point to $0.25c$ for the two last results? if so, how can I achieve that without restarting the simulations and specifying the $0.25c$ as reference point?

Your help is greatly appreciated.

Depending on the target audience, it may be more intuitive to show the movement of the Centre Of Pressure (COP) with different AOAs

Is it OK to compare all three moment coefficient together given that the reference points are different?

No, it's not ok. The pitching moment will be different at different reference locations, unless lift is 0.

Do I have to change the reference point to 0.25c for the two last results? if so, how can I achieve that without restarting the simulations and specifying the 0.25c as reference point?

The industry standard for reporting and comparing pitching moment is quarter chord, because the classic Thin Airfoil Theory concludes that infinitely thin airfoils in subsonic flow have their aerodynamic centres at quarter chord.
That being said, there is absolutely nothing wrong gathering the pitching moment data about any other point, as long as you stay consistent when comparing to other data (or more importantly other people's data). You can convert the pitching moment from one point (say $x_0$) to any other point ($x$) via:
$$M(x) = (L\cos\alpha+D\sin\alpha)(x-x_0) + M_0 \approx L(x-x_0) + M_0$$
where $L$ is the measured lift, $D$ is the measured drag, and $M_0$ is the measured pitching moment at $x_0$. The approximation works great at small AOA ($\alpha$).

Thank you for the detailed answer. What about the sign in $M(x)$ formula? is $x$ the same as $d$ in the figure above?

@izri_zimba x is positive toward TE. M is positive CW.

@izri_zimba You need lift. M_0.25 = (0.25 - 0.75)*L + M_0.75

Yes, thanks, I made a typo. but shouldn't Lift multiplied with something like $\cos(\alpha)$? what about drag contribution?

The airfoil is pitching from -15° to 27°. So I think lift should be multiplied by the cosine of angle of attack. am I wrong?

@izri_zimba This is all based on small angle assumptions (sin(t)~t, cos(t)~1). Drag effect is negligible at small angles (small*small = negligible). Of course, you can always choose to preserve all the nonlinear terms.

Sorry, I couldn't understand, could you please update the formula, my guess is that it should be something like: $$ M(x) = (x-x_0)(L\cos\alpha-D\sin\alpha) + M_0 $$

@izri_zimba Updated. You have incorrect sign on D.

You forgot to multiply by $(x-x_0)$

@izri_zimba Good catch. Thanks.


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Why does a helicopter that uses the Coandă effect need a direct jet thuster aswell
Why does a helicopter that uses the Coandă effect instead of a tail rotor need a direct jet thruster as well. Why couldn't they just vary the pitch of the ducted fan used for the Coandă effect?

The responses to changes in the Coandă effect are too slow to be useful as a precise anti-torque input control operated from the pedals. Remember that the sideways lift is coming from the main rotor downwash. CE is only allowing a tubular boom to make substantial side ways lift from the downwash, not producing any sideways thrust itself from the fan flow.
Hughes found that it only worked if you used CE induced lateral lift from the downwash as a baseline anti-torque force, and used the thruster to fine tune the total force for yaw control.


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How does pulling back on the yoke make you recover from a flat spin
One method I read here, to recover from a flat spin, is to pull back on the yoke or stick, which would push the tail down when facing to the back, and up a bit when facing forward. I assume, if the plane as a whole, acts like a spinning top, it would be like tapping a spinning disc up on the side facing the oncoming air and down on the side facing back. That, with the gyroscopic effect should tip the plane sideways (to the right out of a LH spin) is that correct?

Please see if the following are helpful: https://aviation.stackexchange.com/questions/46935/what-is-the-procedure-to-enter-maintain-and-recover-from-a-flat-spin or https://aviation.stackexchange.com/questions/2170/is-it-possible-to-recover-from-a-flat-spin/2211#2211

@Ralph J Thanks! Rather than tipping the plane, the methods seem to be about coutering the spin by trying to spin the plane the other way, either by the rudder, or by using the ailerons to make the plane act like a windmill, making use of the updraught as the plane falls.


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Is there a benefit or advantage of decreasing the transtion altitude or transition level?
Some counrties have high transition altitudes and transition levels while other countriea have low ones.

Possible duplicate of: https://aviation.stackexchange.com/q/42846/60584

Transition altitudes / levels are instituted so all aircraft have the altimeter setting and will maintain vertical separation.
For aircraft that travel long distances quickly having the ability to set 29.92 (1013) makes it much easier to maintain vertical separation as they are not required to update the altimeter setting every 100 NM or so.
If terrain is a factor, the standard altimeter setting is not accurate enough to avoid conflicts with terrain.
There is a saying "FROM HIGH TO LOW LOOK OUT BELOW".  If flying in an area of low pressure, the pilot who is using 29.92 will be flying closer to the ground than they should.  In this instance, having the correct altimeter setting is vital to main vertical separation from terrain.
Since all aircraft close to terrain needs that new altimeter setting it will also work to maintain vertical separation from aircraft.
The United States has a high transition altitude, and it is above the highest terrain in the country.  It is my opinion, that this would be the reasoning for most countries as well.


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What are dry and saturated lapse rates values or ranges for a rising or lowering parcel of air in degrees Celsius?
Environmental lapse rate average is rounded to 2 degrees celsius per 1000 ft.(Note that environmental lapse rate varies through the day).What about parcel of air lapse rate.? Dry lapse rate is a constant while saturated lapse rate varies based on water vapor in the first place.
Do pilots flying big airlines care about these values?
Apart from being able to calculate cumulus cloud base from Temperature and dew point...
I also heard of something called CAPE seen in Skew T Log P diagram. Do pilots practically flying big planes worry about these? Or is technology different now a days that pilots do not need to care about dry and saturated lapse rate and environmental one...

Yes, they care about weather conditions at various altitudes but do not need to calculate them because the information is already available in the METAR reports.  Reading METARs, weather radar, and other forecasting information is an important part of pre-flight planning.
While dry lapse rates are more or less constant, warmer air has a much higher water carrying capacity than cold air, hence, the moist lapse rate can be much lower because more latent heat is released per parcel of air as more water vapor condenses.  Moist lapse rate is dependent on temperature and relative humidity.


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If you are in a terrain escape maneuver, would you be able to see traffic above you displayed in the cockpit of airliners
nayed
If you are in a terrain escape maneuver, would you be able to see traffic above you in the cockpit of big airliners

What do you think a “terrain escape maneuver” is?  Why do you think (or not think) you’d be able to see traffic above you?

I down voted this question because you really haven’t explained what you mean by “traffic above you”. I think you also need to explain why you think a “terrain escape maneuver” is an important part of seeing  an aircraft above you.

(See chat -- https://chat.stackexchange.com/transcript/message/61400721#61400721 and subsequent (may wrap to "next day"))

Well Yosef, can you help us help you by giving some context?  I assume from your other posts that you mean climbing at Vx in a light civil trainer, but what kind?  Or are you asking about military aircraft?  Because in a max G pull to the vertical you can definitely see above you.  There might be a good question in this, but you need to make it complete or you will get conflicting answers as you have here…

Yes, if you are in an aircraft with good upward visibility, and you tilt your head upwards, it is reasonable to presume that you could see traffic above you within your filed of view as you pitched up.
High wing aircraft generally don't have as good visibility in this direction as low wing aircraft, but a lot depends on the structure of the canopy and other possible obscurations to vision.

No, you will not see aircraft above you. Very few aircraft have adequate visibility that would allow you to see other aircraft directly above. Fighter aircraft with a “bubble” canopy are one exception.

Directly above, and at zero pitch I agree.  But if you crane your head forward, pitch up and look around you can clear the space above to the extent that you ought to be able to avoid a collision.  (not saying it's a very good question though, because it obviously depends on the caveats I have given...)

I just added a “no” answer to counter your “yes” answer which sort of points out that it’s a poor question to begin with.

Yeah, good point...


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Are control surfaces less effective at supersonic speeds?
I am unsure if this is correct but this is my current explanation:
Once past supersonic speeds, the larger the speed the larger the divot of air pressure around the plane. Because these divots can get so large at high speeds, control surfaces are ineffective due to the air not flowing close enough to them. When I say divots, I am referring to the "wavefronts" and "shock cone" around the aircraft. Image for reference (from Wikipedia):

No, in fact rockets and missiles requires small surfaces (see e.g. SpaceX's grid fins which are very effective on high speed). Air and its (static) kinetic energy are the same per volume (but it varies with altitude and temperature).

There may be less longitudinal stability due to (wave) pressure on the nose rather than significant loss of control surface authority.  Pressure differential within the "divot" (being a golfer, I like it) is measured in pounds per square foot (divide by 144 for psi). It may be significant at origin and worth study.  Early high speed rocket aircraft (the X series) had a tendency to lose stability, but these were flown in extremely thin atmosphere.  Loss of stability at higher speeds was known as "divergence".  Notice the build of the SR-71 was extremely stable.

Once past supersonic speeds, the larger the speed the larger the divot of air pressure around the plane.

There are no divots around the plane. Even at supersonic speeds air keeps on following the plane's contour just like at subsonic speeds. When the airplane comes, air just gets out of the way; at subsonic speeds air begins already well in front of the airplane to move away while at supersonic speeds it does it instantaneously via shock and compression waves.
No matter the speed, lift and drag always follow the well known relationships:
$\begin{cases}
L=½ \rho V² SC_l\\
D=½ \rho V² SC_d
\end{cases}$
so that, in general, the bigger the speed the bigger the aerodynamic forces.
Anyway both $C_l$ and $C_d$ also change with the speed. In particular, all the rest being the same:

$C_d$ at supersonic speed is always bigger than at subsonic speed due to the wave drag;
$C_l$ at supersonic speed is lower than at subsonic speed with a ratio that, at least theoretically, for a thin airfoil is 4/2π.

So you're right that

control surfaces effectiveness lowers at supersonic speeds

since $C_d$ is bigger and $C_l$ is lower but this is more than compensate for by the higher speed $V$.

One can measure the supersonic cone with Schlieren photography.  This has been done with rifle bullets.  The angle of the cone can be used to determine Mach number by the formula M = 1/(sine cone angle).

What are the values Cd, cl, and S in these equations?

Coefficient of drag, C. of lift, and area.

$\frac{1}{2} \rho V^2$ may only be used for incompressible dynamic flow, up to about M= 0.5.

@Koyovis: How is the relationship at higher speed?

Would you say control surfaces are still less effective due to the higher coefficients of lift and drag at higher speeds? At these speeds it takes less force from the control surfaces to turn, so couldn't it be argued to be more efficient at higher speeds? Does this make sense?

@SamJones: it really depends on how you want to measure efficiency. In my opinion lower lift and/or higher drag means (all the rest beings the same) less efficiency

The wiki article for dynamic pressure!

@Koyovis: in basic aerodynamic equations (like the one for L, D and M) all the effects due to "numbers" (Reynolds number, Mach number, Strouhal number, ... ) are condensed in the relevant coefficient (of lift, of drag, of moment) and the rest of the equation ($½\rho V²S$) remains the same. This is the beauty of dimensionless coefficients.

$\frac{1}{2} \rho V^2$ is from the Bernoulli equation for incompressible flow. For compressible flow, the equation relevant to Mach number is used - everything changes a lot when the local M = 1 and beyond, right?

@Koyovis: "$½ \rho V²$ is from Bernoulli equation" No, it (and S as well) is from Buckingham theorem and does not change. All the changes are concentrated in the relevant coefficient, which is indeed a function of Re and Ma

And therefore, professionals use the dynamic pressure equation with static pressure and Mach number $p_d = \frac {1}{2} \cdot \gamma \cdot p_{static} \cdot M^2$

@Koyovis: Anderson (but I don't know if he is professional enough for you...) uses the following equations for thin airfoils at subsonic and supersonic speeds respectively:
$\begin{cases}
L=½ \rho V² SC_l\rightarrow C_l=\frac{2\pi\alpha}{\sqrt{1-M²}}\
L=½ \rho V² SC_l\rightarrow C_l=\frac{4\alpha}{\sqrt{M²-1}}
\end{cases}$

I'm familiar with linearised theory for thin airfoils at supersonic velocities.  Please focus on the dynamic pressure equation though, which was what the original comment was about. I'm through with this here.

No.  This is not true.
Control surface effectiveness scales with dynamic pressure.  Faster means more.
Shocks located on a control surface, or on the hinge line can wreak haovc.  However, once you're solidly supersonic, there is no 'divot' in the flow that the aircraft exists in.


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What's the meaning of these jet fighter pilots gestures?
As per title, what's the meaning of the following gestures (if any)?

I expect #3 (both hands up) is to show the ground crew that he is NOT touching the stick or throttle. As the ground crew makes their final prelaunch inspection, Snuffy does not want to get bashed in the head by a moving horizontal stab, or sucked into the engine if he spins it up at the wrong moment.

Please provide the original source for each photo.

ORIGINAL SOURCES:EDITS FROM TIKTOK!!

It looks like some of them are just posing for the photo.


text:
Why do experimental aircraft have such long tubes on the nose?

Why do experimental aircrafts (like the X-35, the Space Shuttle Enterprise, the H160 or the X-15 in the pictures) have such long Pitot tubes at the nose when compared to the production aircraft? Are they for taking test measurements?

Related: https://aviation.stackexchange.com/questions/83113/do-aircraft-designers-gather-real-data-to-confirm-their-design-cfd-wind-tunnel/83115#83115

Yes- you can see an array of little doodads sprouting from the pitot probe in this photo, they are for accurately measuring the direction of airflow in 3 dimensions. They also are long to ensure that the measurements will not be too badly disturbed by the fuselage of the plane itself.

Air pressure varies greatly along the fuselage of the aircraft and while the general average distribution can be modeled quite accurately, local variations can be enough to cause significant error in pressure measurements. Also when aircraft is subjected to extreme flight conditions errors in pitot tubes and static ports can render flight instruments useless.
When test flying new aircraft long pitot tubes and trailing static pressure sensors are used to ensure the probes are free of any interference the body of the aircraft may cause.
During the test flight campaign location of the probes and calibration of flight instruments are finalized so later production aircraft can have accurate pressure sensing within their entire envelope.